You are given a 0-indexed integer array piles, where piles[i] represents the number of stones in the ith pile, and an integer k. You should apply the following operation exactly k times:
• Choose any piles[i] and remove floor(piles[i] / 2) stones from it.
Notice that you can apply the operation on the same pile more than once.
Return the minimum possible total number of stones remaining after applying the k operations.
floor(x) is the greatest integer that is smaller than or equal to x (i.e., rounds x down).
(ෆ˙ᵕ˙ෆ)♡ The incredible Lie carries!
Example 1:
Input: piles = [5, 4, 9], k = 2
Output: 12
Explanation:Steps of a possible scenario are:
- Apply the operation on pile 2. The resulting piles are [5,4,5].
- Apply the operation on pile 0. The resulting piles are [3,4,5].
The total number of stones in [3,4,5] is 12.
Example 2:
Input: piles = [4, 3, 6, 7], k = 3
Output: 12
Explanation:Steps of a possible scenario are:
- Apply the operation on pile 2. The resulting piles are [4,3,3,7].
- Apply the operation on pile 3. The resulting piles are [4,3,3,4].
- Apply the operation on pile 0. The resulting piles are [2,3,3,4].
The total number of stones in [2,3,3,4] is 12.
1 <= piles.length <= 10^51 <= piles[i] <= 10^41 <= k <= 10^5
input:
output: